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How to find a complete way to the folder, end up and copy the contents

jeanid last edited by

So, in the folder on. C:\Users\Server\AppData\Roaming\Spotify\Users There's a lot of folders, but I need to know the name of all folders with the end--user, that's what they call me. eg9no0brqmzrmfy6crh7e7fsy-user and rplc59ty0gmhfy4q8tgaz8gyt-user I'm done. -user I need to know the name of the folder, and copy her contents into another folder.
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simrah

It's through one os library. Yeah, I know that os.system is not a very good tone, but piton work with folding is terrible. So it's better and clearer. The folder is here for Linux. Windows is a little different.

import os
BASE_DIR = '/Ваша папка/'

TARGET_DIR = '/Ваша папка назначения/'
all_dirs = [x[1] for x in os.walk(BASE_DIR)][0]
user_dirs = list(filter(lambda x: x.endswith('-user'), all_dirs))

user_dirs = list((BASE_DIR + x) for x in user_dirs)
Для Linux
for dir_ in user_dirs:

os.system(f'cp -r {dir_} {TARGET_DIR}')
Для Windows
for dir_ in user_dirs:

os.system(f'xcopy {dir_} {TARGET_DIR}')

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Recurring function to find a way in the Lists
Software Programming • c sharp list language curriculum • • obi

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You can use the class, but you don't. The challenge is to circumvent the count. Where the dictionaries are the top of the count. And access keys are ribs. I'll notice it's an orgraph.Apologies for the introduction of your own data entry system, writes to test workability. I wrote it, not a sin, and I told you.There are two numbers on the entrance, the N is the keys and the M is the keys. The following M rows describe the keys in the format: 1st number, the keys following the gap, the keys of other dictionaries. The programme further requires two numbers of X and Y. What dictionary needs to find a way to another.The program is making the shortest way, if it exists, or "Don't exist." string[] Inp = Console.ReadLine().Split(); // Заносим данные ==>> int N = Convert.ToInt32(Inp[0]), M = Convert.ToInt32(Inp[1]); Dictionary<int, List<int>> dict = new Dictionary<int, List<int>>(); for (int i = 0; i < N; i++) dict[i] = new List<int>(); for (int i = 0; i < M; i++) { Inp = Console.ReadLine().Split(); int Temp = Convert.ToInt32(Inp[0]) - 1; for (int j = 1; j < Inp.Length; j++) { dict[Temp].Add(Convert.ToInt32(Inp[j]) - 1); } } Inp = Console.ReadLine().Split(); int X = Convert.ToInt32(Inp[0]) - 1, Y = Convert.ToInt32(Inp[1]) - 1; // <<== Все еще заносим Queue<int> Work = new Queue<int>(); // BFS работает через очередь. bool[] Mark = new bool[N]; // Массив, в котором будем помечать посещенные словари int[] Log = new int[N]; // Потребуется для вывода найденного пути bool Exist = false; // Существует ли наш путь вообще Work.Enqueue(X); // см. Алгоритм BFS (обход в глубину) Mark[X] = true; Log[X] = -1; while (Work.Count > 0) { int v = Work.Dequeue(); for (int i = 0; i < dict[v].Count; i++) { if (!Mark[dict[v][i]]) { if (dict[v][i] == Y) Exist = true; Work.Enqueue(dict[v][i]); Mark[dict[v][i]] = true; Log[dict[v][i]] = v; } } } if (Exist) { List<int> output = new List<int>(); // Восстанавливаем путь for (int v = Y; v != -1; v = Log[v]) output.Add(v); output.Reverse(); for (int i = 0; i < output.Count; i++) Console.Write(output[i] + 1 + " "); Console.WriteLine(); } else Console.WriteLine("Don't exist"); This is an example of work:Input:100 6 1 2 3 4 5 5 2 3 1 45 42 13 54 42 13 3 45 42 54 1 13 Output data:1 3 45 13 I hope I did.
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Not ends asynchronous-completed Task Force (Gulp 4)
Software Programming • javascript node.js gulp task • • guilherme

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Could've done it all by himself.(1) In front of everyone. gulp.src.. Get up. return♪ Got it. return gulp.src (Thanks to comrade https://ru.stackoverflow.com/users/184903/bleser )(2) Changed the announcement, and instead of Function Expression, I declare gulp.serieswithin which I declare gulp.parallel:gulp.task('dev:build', gulp:series( gulp.parallel('dev:html', 'dev:css', 'dev:js') ));
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Django Template - Get the variable from URL
Software Programming • python python 3.x django • • amarantha

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{{ request.resolver_match.kwargs.id }}
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Add panels to a window with Pyqt
Software Programming • python python 3.x pyqt5 graphic languageal interf qmainwindow • • Amritpal

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You have the following mistakes:In the connection with the signal triggered You must pass the name of the function, not the function evaluated, the right thing is: fileMenu.triggered.connect(self.buscarProductos).widgets does not have the addWidget() method so it throws the error.On the other hand before writing it is better to have a scheme of what is expected to have, so the structure should be the following:MainWindow └── centralwidget └── tabwidget ├── GeneralTab ├── GeneralTab ├── ... └── GeneralTab Then as we see only there must be a QTabWidget and in the FindProduct CodeScreen you are creating others(on the other hand SearchProductsScreen is not a widget so tabdialog.show() It won't work), but I find it unnecessary to create that class so I will remove it to avoid confusion.Using the above we get the following:class MainWindow(QMainWindow): def __init__(self): super().__init__() self.initUI() def initUI(self): centralwidget = QWidget() self.setCentralWidget(centralwidget) lay = QVBoxLayout(centralwidget) menubar = self.menuBar() fileMenu = menubar.addMenu('Buscar...') impMenu = QAction('Productos', self) fileMenu.addAction(impMenu) fileMenu.triggered.connect(self.buscarProductos) self.tabWidget = QTabWidget() lay.addWidget(self.tabWidget) self.setWindowTitle('IA - Análisis datos') self.show() def buscarProductos(self): #limpiarPantalla() filename = "." fileInfo = QFileInfo(filename) self.tabWidget.addTab(GeneralTab(fileInfo), "General")
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How to separate the text from the list element
Software Programming • python python 3.x • • jules

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lst = ["/view_profile?pet_id=34532234", "/view_profile?human_id=546476346"] new = [tuple(i.split('=')) for i in lst] # [('/view_profile?pet_id', '34532234'), ('/view_profile?human_id', '546476346')] # либо генератор new = (tuple(i.split('=')) for i in lst) # и pet_id :) new = [i.split('=')[1] for i in lst if 'pet_id' in i]
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How to find the end and start the line in TextBox.
Software Programming • visual • • Jolied

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Field SelectionStart is the starting point. SelectionLength is the length of the fragment
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Delete folder and contents with php
Software Programming • php • • Pearlaqua

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You can use this PHP function that erases the folder and its internal files: function delTree($dir) { $files = array_diff(scandir($dir), array('.','..')); foreach ($files as $file) { (is_dir("$dir/$file")) ? delTree("$dir/$file") : unlink("$dir/$file"); } return rmdir($dir); } delTree('caminho/da/pasta/aqui'); This function is in itself http://www.php.net/manual/en/function.rmdir.php#110489 I hope I helped.
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How to copy the contents of two files into a third file in c?
Software Programming • c language file • • Pearlaqua

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So I'm a beginner in programming and I'm learning to program in C. A few days ago I learned about files (in fact I'm still learning). Your code working properly is this:#include <stdio.h> #include <stdlib.h> int main() { FILE *arqum; FILE *arqdois; FILE *arqtres; char ch,ch1,leitor[100],leitor2[100]; char nome1[20],nome2[20]; printf("\nDigite o nome primeiro arquivo:\n"); gets(nome1); arqum=fopen(nome1, "r"); //abre o arquivo para leitura if(NULL==arqum) { printf("O arquivo não pode ser aberto. \n" ); system("Pause"); exit (1); } ch=fgetc(arqum); while(ch!=EOF) { putchar(ch); ch=fgetc(arqum); } fclose (arqum); printf("\nDigite o nome do segundo arquivo:\n"); gets(nome2); arqdois=fopen(nome2, "r"); //abre o arquivo para leitura if(NULL==arqum) { printf("O arquivo não pode ser aberto. \n" ); system("Pause"); exit (1); } ch1=fgetc(arqum); while(ch1!=EOF) { putchar(ch1); ch1=fgetc(arqdois); } fclose (arqdois); arqtres=fopen("arquivotres.txt","a+"); if(NULL==arqtres) { printf("O arquivo não pode ser aberto. \n" ); system("Pause"); exit (1); } arqum=fopen(nome1, "r"); //abre o arquivo para leitura DE NOVO if(NULL==arqum) { printf("O arquivo não pode ser aberto. \n" ); system("Pause"); exit (1); } while(fgets(leitor,100,arqum)!=NULL); fputs(leitor,arqtres); fclose(arqtres); fclose (arqum); // fecha arquivo um arqdois=fopen(nome2, "r"); //abre o arquivo para leitura DE NOVO if(NULL==arqudois) { printf("O arquivo não pode ser aberto. \n" ); system("Pause"); exit (1); } arqtres=fopen("arquivotres.txt","a+"); while(fgets(leitor2,100,arqdois)!=NULL); fputs(leitor2,arqtres); fclose (arqdois); // fecha arquivo dois fclose(arqtres); return 0; } The solution to your problem was that arqum and arqdoiswere closed at the time you were passing leitor and leitor2to the arqtres, that is, you were using the arqum and arqdois to open the respective files requested by the user. In this way, they served as pointers to the files that the user requested to read and and and as you closed them, did not make sure the passage of the strings read to the arqtres. Then open them (I do not know if this is the correct syntax of the Portuguese kkk) again. You can see on the linesarqum=fopen(nome1, "r"); //abre o arquivo para leitura DE NOVO andarqdois=fopen(nome2, "r"); //abre o arquivo para leitura DE NOVO and then I was closing after use. That's it. I hope you understand.
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How to use a function within another function in python3?
Software Programming • python python 3. x • • rielyn

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If in function nome() you returns the value read, you will need to store that value in a variable. The variable name that you defined within nome() does not exist within intro(). They're different scopes.def nome(): name = input('Poderia nos dizer seu nome? ') return name def intro(): name = nome() print('Olá {}, seja bem vindo!'.format(name)) intro() Understand that I removed str() the input also, because this is redundant; the return of input will always be a string.
Find files in a folder with PHP
Software Programming • php html imagen • • irl

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You can do the following// Ruta del directorio donde están los archivos $path = '/tmp'; // Arreglo con todos los nombres de los archivos $files = array_diff(scandir($path), array('.', '..')); Then you go through the arrangement and make it simple. explode to each element// Obtienes tu variable mediante GET $code = $_GET['codigo']; foreach($files as $file){ // Divides en dos el nombre de tu archivo utilizando el . $data = explode(".", $file); // Nombre del archivo $fileName = $data[0]; // Extensión del archivo $fileExtension = $data[1]; if($code == $fileName){ echo $fileName; // Realizamos un break para que el ciclo se interrumpa break; } }
C

Error: 'int' object is not subscriptable
Software Programming • python list • • courtlanda

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As I said, the error is given in this line why you did: [i[j]]. With this code, you are accessing the position j of the variable i; how i is a type variable int, it is not possible to do this, generating the error. I believe the desired was to do [i][j], to access the position j of value in position i.With this change your code seems to work as desired.See working on https://repl.it/Kv6x .However, I believe the code can be improved. Motif that of me answer and not only vote to close the question as typing error. Your code has many addictions that are common to people starting in Python already knowing another language - usually C. This gets very clear when you use the range(len(...)). Its code is functional, but not idiomatic, that is, it does not use the proper tools that language offers to solve the problem.First, I would not store the name and distance values in separate lists. They are given directly related to each other and makes sense to store them together. I would use a dictionary with the keys nome and saltos to store the information and store that dictionary in a list. Something like:atletas = [] while True: nome = input("Nome?") if not nome: break saltos = [] for i in range(5): salto = float(input("Distância {}?".format(i+1))) saltos.append(salto) atletas.append({ "nome": nome, "saltos": saltos }) for atleta in atletas: print("Nome:", atleta["nome"]) print("Saltos:", atleta["saltos"]) print("Média:", sum(atleta["saltos"])/5) Note that average calculation is much simpler if you use sum(...)/5. When reading the name, you do not need to convert to string, for the return of input It'll always be this guy right now. In fact, the program does not talk about the need to store these values in a list, so the display of the name and average could be done in the same loop as the reading, not needing to create the list atletas.See working on https://repl.it/Kva6/0 .
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Ways to get the src files in a local folder with js?
Software Programming • javascript • • amarantha

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I don't know if you've already solved it, but in case the doubts here are an example of use or what you can do to create links to your resources when you don't know the route they come from.To say that URL.createObjectURL() Like URL.revokeObjectURL are synchronous methods and are also instantaneous methods, so they become quite useful, then we continue with the example:window.onload = ()=>{ //Obtenemos los elementos HTML. const inputFile = document.getElementById("inputFile"); const audioSources = document.getElementById("audioSources"); //Aqui tendras tus rutas disponibles (array de rutas de URL.createObjectURL). let fileUrls = []; function handleFiles(ev){ //Reiniciamos cada vez para no guardar audios que no se usan. //No reinicies el array si no es el comportamiento que deseas. //Tampoco reinicies audioSources si no lo deseas. fileUrls = []; audioSources.innerHTML = ""; //Obtenemos los archivos como un array (originalmente es un FileList). const files = Array.from(ev.currentTarget.files); //Podemos usar forEach puesto que hicimos Array.from... files.forEach((file, i) => { //Creamos un nuevo elemento audio para añadirlo posteriormente y ademas //Añadimos la nueva URL del recurso a la lista de URLS. const newSource = document.createElement("audio"); fileUrls.push(URL.createObjectURL(file)); //Cuando haya terminado de cargar podemos usar URL.revokeObjectURL //El audio seguira funcionando, debido a que ya cargo, configuramos tambien. //Las propiedades del audio para poder hacer uso de sus controles y que carguen //nuestro recurso. newSource.oncanplaytrough = ()=> URL.revokeObjectURL(fileUrls[i]); newSource.src = fileUrls[i]; newSource.controls = "controls"; //Agregamos los audios a nuestro contenedor de audios y adicionalmente ponemos un //Salto de linea al final. audioSources.appendChild(newSource); audioSources.appendChild(document.createElement("br")); }); } //Se ejecuta cada vez que los archivos seleccionados cambian. inputFile.addEventListener("change", handleFiles); }<input type = "file" accept = "audio/*" id = "inputFile" multiple> <div id = "audioSources"> </div>This is a way to use URLs of resources to which you do not have access to your route.
to make comments on the post at django
Software Programming • python django • • Bogopo

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First, we need to create a simple model:# models.py from django.db import models class Name(models.Model): name = models.CharField(max_length=100) Make migration:python manage.py makemigrations python manage.py migrate It's easier to use. https://docs.djangoproject.com/en/2.1/topics/forms/modelforms/ which is needed to create models# forms.py from django.forms import ModelForm from .models import Name class NameForm(ModelForm): class Meta: model = Name fields = ('name',) During the POST request, retain the name in the database, redesign on the same page, extract the data from the OBD during the GET of the request and display them on the page.# views.py from django.http import HttpResponseRedirect from django.shortcuts import render from .forms import NameForm from .models import Name def get_name(request): if request.method == 'POST': form = NameForm(request.POST) if form.is_valid(): # Сохранение формы form.save() # Редирект на ту же страницу return HttpResponseRedirect(request.path_info) else: # метод GET form = NameForm() # Получение всех имен из БД. names = Name.objects.all() # И добавляем names в контекст, чтобы плучить к ним доступ в шаблоне return render(request, 'name.html', {'form': form, 'names': names}) Shablon name.html<form action="" method="post"> {% csrf_token %} {% if names %} {% for name in names %} {{ name.name }}<br> {% endfor %} {% endif %} {{ form }} <input type="submit" value="Submit"> </form>
How do you make the contents of the section fully complete?
Software Programming • html cs • • briley

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Why do you need style?section { padding: 60px 0 !important; } If the content doesn't fall under the menu, it's better to use it, like I'm catching, to the class element. .wrapperFor class .services-first-slider Add the style.background-size: cover; Then the picture will be all over the block.
C

When copying the file, a folder is being created
Software Programming • php • • chanisef

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If the directory is established, the path has been put to the argument of the function. According to your code, there were two options. First, seal mkdir("1/$v") instead $dsecond - cycle < count($t)instead < count($t) - 1So the name of the file was also in the name of the file. How did we find out the problem was at -1However, in order to meet the challenges, try to use suitable tools. For example, there is no need to deal with ways on its own, so there are functions. There is also no need to write many of the conditions imposed, etc. Maybe it's better to just do some data in a few ways and then just process them. Also, constant concentration and calculation of the route increases the risk of the seals and reduces the clarity and readability of the code (not to mention that when the path is changed, each line will have to be ruled) and the code should be simple as vain, the easier the better.This is the example of the code (possibly with seals) where these issues are not so relevant:$files = [ "/uploads/2017/04/provoloka.jpg", "/uploads/2017/02/provoloka-vyazalnaya.jpg" ]; // из какой директрию и в какую копируем $sourceDir = dirname(FILE); $targetDir = $sourceDir."1" ; // пути для файлов $data = array_map(function($f) use ($sourceDir, $targetDir){ return [ 'from' => $sourceDir.$f, 'to' => $targetDir.$f, ]; }, $files); // фильтруем только то, что действительно следует копировать $data = array_filter($data, function($c){ return file_exists($c['from']) && !file_exists($c['to']); }); // работаем с файлами foreach($data as $c){ $d = dirname($c['to']); if(!file_exists($d)){ mkdir($d, 777, true); } copy($c['from'], $c['to']); } for special aesthets, you can do $data = array_column($data, 'to', 'from') and type cycle foreach($data as $from => $to){ ... }

How to find a complete way to the folder, end up and copy the contents

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